CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Omitting tau can move an exact fine root

Statement refuted

The claim that replacing the tau-corrected coarse right side by the restricted fine right side $R f_{h}$ always preserves an exact fine-grid root is false.

Facts & Assumptions

Given: The two-dimensional fine residual $F_{h} (u) = A u$ with $A = (\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array})$ , right side $f_{h} = A (1, 0)^{T}$ , prolongation $P = (1, 1)^{T}$ , iterate restriction $I u = (u_{1} + u_{2}) / 2$ , residual restriction $R = P^{T}$ , and rediscretized coarse residual $F_{H} (v) = 3 v$ .

[F1]

Finite-budget nonlinear FAS two-grid cycle: the FAS coarse right side includes $F_{H} (I u) + R (f_{h} - F_{h} (u))$ and the prolongated update is $u + P (v - I u)$ .

[F2]

An exact fine root is fixed by the FAS V/W cycle: with the tau-corrected right side and the selected zero-defect coarse root, an exact fine root is fixed.

Counterexample

1.1givenalgebra

Set $u_{*} = (1, 0)^{T}$ . Then $f_{h} = A u_{*} = (2, - 1)^{T}$ , so $u_{*}$ is an exact fine root. Its restriction is $I u_{*} = 1 / 2$ and the restricted right side is $R f_{h} = 2 - 1 = 1$ .

2.1F1step 1.1algebra

With tau omitted, the coarse equation is $3 v = R f_{h} = 1$ , whose exact solution is $v = 1 / 3$ . The fine correction is $u_{*} + P (v - I u_{*}) = (1, 0)^{T} + (1, 1)^{T} (1 / 3 - 1 / 2) = (5 / 6, - 1 / 6)^{T},$ which is not $u_{*}$ . Thus the false claim fails for this SPD fine system and this specified rediscretized coarse operator.

3.1F1F2step 1.1algebra∎

For the same inputs, the correct tau is $F_{H} (I u_{*}) - R f_{h} = 3 / 2 - 1 = 1 / 2$ , so the tau-corrected coarse right side is $3 / 2$ and its selected root is $v = 1 / 2 = I u_{*}$ . The resulting correction is zero, in accord with [F2].

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